Question

The vapor pressure of ethanol is 100 mmHg at 34.9°C. What is its vapor pressure at 55.5°C?

Answer 1

use the clausius clapeyron equation...

ln (P1 / P2) = ΔHvap / R x (1 / T2 - 1 / T1)

P1 = ???

P2 = 100 mmHg

ΔHvap = 39.3x10^3 J/mole

R = 8.314 J/moleK

T2 = 34.9 C = 34.9+273.15 K = 308.05 K

T1 = 71.5 C = 71.5+273.15 K = 344.65 K

plugging in...

ln (P1 / 100 mmHg) = [(39300 J/mole) / (8.314 J/moleK)] x (1 / 308.05 K - 1 / 344.65 K)

ln (P1 / 100 mmHg) = 4227K x (3.447x10^-4 / K)

ln (P1 / 100 mmHg) = 1.630

e^ln (P1 / 100 mmHg) = e^(1.549)

P1 = 100 mmHg x 5.102 = 510 mmHg... 3 sig figs